Question

PT is a bisector of angle QPR in triangle QPR and PS parallel to QR find the value of
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Answer 1

Answer:

ur answer is 80° Hope this helps u

Answer 2

Answer:

In∆PQS,<QPS=180°-(50°+90°)=180°-140°=40°

<QPT=<RPT

<QPR=180°-(50°+30°)=180°-80°=100°

<QPR=<QPT+<RPT=2<QPT

100°/2=<QPT

50°=<QPT

x=<QPT-<QPS=50°-40°=10°